Let $R$ be a ring, $X,Y,Z$ and $T$ four $R-$ modules such that there exists a short exact sequence $$0 \rightarrow X \xrightarrow{f_1} Y \xrightarrow{f_2} Z \xrightarrow{f_3} T \rightarrow 0 $$
Prove the following:
(i) If $X,Y$ and $T$ are Noetherian modules, then $Z$ is Noetherian.
(ii) If $X,Z$ and $T$ are Noetherian modules, then $Y$ is Noetherian.
I couldn't prove any of them, I assume both parts have similar proofs so I'll write what I've tried to do in (i) and I would appreciate some help with one of the two so I can complete the rest by myself.
In (i), there are alternative ways to show that $Z$ is Noetherian. I chose to show that every ascending chain is stationary.
So let $(Z_k)_{k \in \mathbb N}$ be an ascending chain in $Z$. Then $(f_3(Z_k))_{k \in \mathbb N}$ is an ascending chain in T. The same for $({f_2}^{-1}(Z_k))_{k \in \mathbb N}$ in Y. By hypothesis both of these chains are stationary. Let $N \in \mathbb N$ be such that $f_3(Z_n)=f_3(Z_N)$ for all $n \geq N$ and ${f_2}^{-1}(Z_n)={f_2}^{-1}(Z_N)$ for all $n \geq N$.
Take $x \in Z_n$ for $n \geq N$, I would like to show $x \in Z_N$. We have $f(x) \in f_3(Z_n)=f_3(Z_N)$, so there is $y \in Z_N$ such that $f_3(y)=f_3(x)$. But then $f_3(y-x)=0$, so $x-y \in Ker(f_3)=Im(f_2)$. Here I got stuck, there is $w \in Y$ such that $f_2(w)=x-y$, but in which submodule of $Y$ is $w$? And I don't see where to use that $X$ is also Noetherian. Any help would be greatly appreciated.
I suppose that you know that if we have a s.e.s. $0\to M'\to M\to M''\to 0$, then $M$ is noetherian iff $M'$ and $M''$ are noetherian. (If not, take a look here.)
Now split your exact sequence in two s.e.s.: $0\to X\to Y\to K\to 0$ and $0\to K\to Z\to T\to 0$.
(i) $Y$ noetherian $\Rightarrow$ $K$ noetherian...
(ii) $Z$ noetherian $\Rightarrow$ $K$ noetherian...